ch13-p036 - 36. (a) From Eq. 13-28, we see that v0 = GM / 2...

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36. (a) From Eq. 13-28, we see that 0 /2 E vG M R = in this problem. Using energy conservation, we have 1 2 mv o 2 GMm/R E = – GMm/r which yields r = 4 R E /3. So the multiple of R E is 4/3 or 1.33. (b) Using the equation in the textbook immediately preceding Eq. 13-28, we see that in this problem we have K i = GMm/ 2 R E , and the above manipulation (using energy
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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